Theorem freq2 4268

Description: Equality theorem for the well-founded predicate. (Contributed by NM, 3-Apr-1994.)

Assertion

Ref	Expression
freq2	⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

Proof of Theorem freq2

Dummy variable 𝑠 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		frforeq2 4267	. . 3 ⊢ (𝐴 = 𝐵 → ( FrFor 𝑅𝐴𝑠 ↔ FrFor 𝑅𝐵𝑠))
2	1	albidv 1796	. 2 ⊢ (𝐴 = 𝐵 → (∀𝑠 FrFor 𝑅𝐴𝑠 ↔ ∀𝑠 FrFor 𝑅𝐵𝑠))
3		df-frind 4254	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑅𝐴𝑠)
4		df-frind 4254	. 2 ⊢ (𝑅 Fr 𝐵 ↔ ∀𝑠 FrFor 𝑅𝐵𝑠)
5	2, 3, 4	3bitr4g 222	1 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

Syntax hints:   → wi 4   ↔ wb 104  ∀wal 1329   = wceq 1331   FrFor wfrfor 4249   Fr wfr 4250

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-ral 2421  df-in 3077  df-ss 3084  df-frfor 4253  df-frind 4254

This theorem is referenced by:  weeq2  4279